For problem 2 I think a) [i] P(one lost on A,one lost on B)=2!(1/5)(1-1/5)(1/16) [ii] P(both lost on B)=[(1-1/5)^2]*(1/16) [iii] P(both lost on A)=(1/5)^2
I don't really understand the solutiong they give, but i dont really think the solution for a)[ii] is correct. The P for one case lost in B is 1/16, how can lost both in B even higher than lost one?
